In part (a), which radii are used in ln(r3/r2) for the insulation layer?

r2 = 0.0275 m (outer pipe/inner insulation) and r3 = outer insulation radius

After using Q/L = 2πk(T2 - T3)/ln(r3/r2), what is the correct step to find r3?

ln(r3/r2) = 2πk(T2 - T3)/(Q/L) = 0.0419, so r3/r2 = e^0.0419 = 1.0428

ln(r3/r2) = (Q/L)/[2πk(T2 - T3)]

Why is the inside pipe surface temperature only about 300.46 °C, barely above 300 °C?

Because the pipe has high conductivity (20 W/m°C) and thin wall, so ln(r2/r1)/(2πk_p) is tiny and the drop is only 0.455 K

Because the insulation raises the pipe temperature

Because the heat flow through the pipe is zero

Because convection inside the pipe removes the heat

2071 Magh · 6 marks

Cast iron pipe with 85% magnesia insulation

Short answer

Using the cylindrical conduction equation Q/L = 2πk (T₂−T₃)/ln(r₃/r₂) with 600 W/m gives a minimum insulation thickness of 1.18 mm. Applying the same heat flow through the pipe wall yields an inner surface temperature of 300.46 °C.

A standard cast iron pipe inner diameter 50 mm and 2.5 mm thick is insulated with 85 percent magnesium insulation, K = 0.02 W/m°C. Temperature at the interface between the pipe and the insulation is 300°C. The allowable heat loss through the pipe is 600 W/m length of pipe and for the safety; the temperature of the outside surface of insulation must not exceed 100°C. Determine a) Minimum thickness of insulation required and b) The temperature of inside surface of the pipe assuming its thermal conductivity 20 W/m°C.

Given: pipe inner diameter 50 mm so r₁ = 0.025 m; wall thickness 2.5 mm so r₂ = 0.0275 m; insulation k = 0.02 W/m·°C; temperature at the pipe-insulation interface T₂ = 300 °C; allowable heat loss Q/L = 600 W/m; outside surface of insulation T₃ ≤ 100 °C; pipe conductivity k_p = 20 W/m·°C.

(a) Minimum insulation thickness

Conduction through the insulation (cylindrical layer, per metre):

Q/L = 2πk (T₂ − T₃)/ln(r₃/r₂)

ln(r₃/r₂) = 2πk (T₂ − T₃)/(Q/L) = 2π × 0.02 × (300 − 100)/600 = 0.04189

r₃/r₂ = e^0.04189 = 1.04278

r₃ = 0.0275 × 1.04278 = 0.02868 m

Thickness = r₃ − r₂ = 0.02868 − 0.0275 = 0.00118 m

Minimum insulation thickness ≈ 1.18 mm (the heat loss limit of 600 W/m is a large allowance, so a very thin layer meets it; a thicker layer would give a lower loss and a lower surface temperature).

(b) Inside surface temperature of the pipe

Conduction through the pipe wall carries the same 600 W/m:

Q/L = 2πk_p (T₁ − T₂)/ln(r₂/r₁)

T₁ − T₂ = (Q/L) ln(r₂/r₁)/(2πk_p) = 600 × ln(0.0275/0.025)/(2π × 20) = 600 × 0.09531/125.66 = 0.455 K

T₁ = 300 + 0.46 = 300.46 °C (the pipe wall adds almost no temperature drop).

✓ Answer checked by Claude Opus 5.5 (high effort)

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Cast iron pipe with 85% magnesia insulation

(a) Minimum insulation thickness

(b) Inside surface temperature of the pipe

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